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20 February, 01:15

A circuit consists of a 12.0-V battery connected to three resistors (44ohm, 17ohm and 110ohm) in series. a. Find the current that flows through the battery. I=__mAb. Find the potential difference across the 44ohm resistor. V1=__Vc. Find the potential difference across the 17ohm resistor. V2=__Vd. Find the potential difference across the 110ohm resistor. V3=__V

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  1. 20 February, 01:47
    0
    a) I = 0.0702A = 70.2mA

    b) V1 = 3.09V

    c) V2 = 1.19V

    d) V3 = 7.72V

    Explanation:

    Given that the circuit consist of three series resistors.

    For resistors arranged in series, the total resistance R can be given as:

    R = R1 + R2 + R3

    R1 = 44 ohms

    R2 = 17 ohms

    R3 = 110 ohms.

    R = 44 + 17 + 110 = 171 ohms

    V = 12 V

    a) The current of a circuit is given by;

    Potential difference V = current * total resistance

    V = IR

    Making I the subject of formula,

    I = V/R

    I = 12/171 = 0.0702A

    I = 7.02mA

    b) the potential difference across any resistors is given by:

    V = IR

    Since the arrangement is parallel, the same current flows through each of the resistors.

    V1 = IR1

    V1 = 0.0702 * 44

    V1 = 3.09V

    c) applying the same rule as b above:

    V2 = IR2

    V2 = 0.0702 * 17

    V2 = 1.19V

    d) applying the same rule as b above.

    V3 = IR3

    V3 = 0.0702 * 110

    V3 = 7.72V
  2. 20 February, 02:22
    0
    I = 70.2mA

    V1 = 3.09V

    V2 = 1.19V

    V3 = 7.72V

    Explanation:

    Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

    From ohm's law

    Voltage (V) = current (I) * resistance (R)

    I = V/R = 12/171 = 0.0702A = 0.0702*1000mA = 70.2mA

    V1 = I*R1 = 0.0702*44 = 3.09V

    V2 = I*R2 = 0.0702*17 = 1.19V

    V3 = I*R3 = 0.0702*110 = 7.72V
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